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Q.
Atomic weight of Boron is 10.81 and it has two isotopes $_5B^{10}$ and $ _5B^{11}$ Then the ratio of $_5B^{10} :_5B^{11}$ in nature would be
Atoms
Solution:
Let ${ }_{5} B^{10}$ be present as $x \%$
and percentage of ${ }_{5} B^{10}=(100- x )$
$\therefore$ Average atomic coefficient $=\frac{10 x+11(100-x)}{100}$
$=10.81 \Rightarrow x=19$
$\therefore \%$ of ${ }_{5} B^{11}$ is $100-19=81$.
Ratio is $19: 81$